I would like to wish everyone, including all haters and losers (of which, sadly, there are many) a truly happy and enjoyable Memorial Day!
— Donald J. Trump (@realDonaldTrump) May 24, 2015My thesis, which I defended last Monday, includes a big list of acknowledgements. Unlike the very stately, presidential message above, it does not include nods to haters or losers. It does include a few unusual entities, like the United States and the Coca-Cola Foundation, but everything remains positive.
Even taking a pretty expansive view of who deserved an acknowledgement, I still made some omissions for space or style. Below is a list of entities, thanked without constraints.
Service workers. Everyday, something in my life is easier because of cleaning staff, restaurant workers, delivery drivers, or other service workers. These men and women work hard, menial jobs and often get very little thanks or recognition (even less than administrative staff like secretaries). For example, the Rutgers math department cleaning staff often works at night, when you can’t see them! Thank you to all the workers who make life a thousand times easier for all of us.
Euclid et al. Although “mathematics” consists of many people working constantly to push our boundary of knowledge forward, there are some people that pushed this boundary further than anyone else. The mathematical community owes a great debt to these legendary mathematicians of yore. Some of the “standard” mentions are Euler, Gauss, Newton, and Leibniz, but I feel that a special nod must be given to Euclid for being the first to shape mathematics into its modern form.
Taxpayers. About 66% of all academics work at public universities. Even outside of public universities, many grants are awarded by NSF, NSA, or other federal agencies, and students often pay tuition using federal loans. Maybe Princeton and Harvard could get by with private money, but the majority of academics work at places that can’t. It is thanks to the United States and her citizens that I and most academics have work. A grateful thank you to them.
Tech wizards. It is unthinkable to do mathematics without the computer tools I use every day. Leslie Lamport and Don Knuth made $\LaTeX$, the symbolic computation group at the University of Waterloo created Maple, and Guido van Rossum designed the almost-perfect Python. I must give a special thank you to the late Bram Moolenaar, who created the truly stupendous text editor Vim. I do 90% of my work by writing notes in Vim, and that work is all the more pleasant and efficient thanks to him.
That covers just about everyone. At this point, if you haven’t been thanked, I have some bad news for you. If it’s any consolation, I’m not dying, so you still have a chance to do something really nice to make the next list. Better start now!
]]>If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many also fall in continued proportion between the numbers which have the same ratios with the original numbers.
If you have trouble seeing $\sqrt{2}$ here, don’t worry; so does everyone who reads the Elements. (Professor David Joyce, explaining how to parse this, remarks “If 1 is considered to be a number, the argument simplifies.”) With all due respect to Euclid, in hindsight we know that mediating all of mathematics through lines and circles makes some things more difficult than they need to be.
Things are better today now that we have symbolic notation, though on occasion people fall into a Euclid-style trap and say that mathematics is symbol crunching. If you can move the right funny letters around in calculus, linear algebra, or differential equations, then you’re a “good student.” But of course math is not just symbol crunching, just like it isn’t just lines and circles.
This historical outline guides me when thinking about AI generated art. There has been a lot of uproar over the past few years from concerned artists about AI tools like Midjourney. Are these tools going to take jobs from already-struggling artists? Are they just stealing original work and repackaging it with fancy post-processing? And whatever you think, can’t we agree that “AI art” is not real art?
This sentiment that “AI art is not real art” seems mistaken. I understand that creating art is very difficult, in the same way that drawing and reasoning about complicated geometric diagrams is difficult. But are the techniques used to create art today what art is? Can it really be said that if you aren’t drawing lines on a tablet, or putting paint on a canvas, you aren’t making art? I don’t buy it.
I suspect that artists lashing out at AI art are more concerned about job security than they are artistic ideals. And fair enough. AI art might not be taking their jobs now, but it probably will. Just look at Cosmo’s cover from 2022, or read the story of Kelly McKernan, whose distinctive art style was reproduced by an AI startup that McKernan is suing for copyright infringement. It is demoralizing to devote a chunk of your life to something only for a newfangled machine to do it better. And yet, like the old folk hero John Henry, you can fight progress your whole life and accomplish nothing but dying.
AI art does not need to be the end of human art. If we, as a society, think that human art is important, then we can fund it. We already do fund it in various ways, like through the National Endowment for the Arts1. I would be happy to see more money go that way. We fund lots of things that no one would ever pay for on their own2, and art is one of the more noble things we could pick.
It’s great that I can ask ChatGPT to explain things, and it’s great that more people can make art using tools like Midjourney. If you are worried about a future where these tools are so widespread that AI does everything, a future where your economic opportunities are extremely limited, a future where humanity is left to passively observe a digital world, then you better start writing blog posts saying nice things about ChatGPT before it takes over.
NEA has a very neat map where you can see how much money they put into individual states. ↩
Like most of academia. Does New Jersey need 100 math Ph.D students at its flagship university? No, of course not. But it’s a nice, fairly cheap thing that brings in smart, motivated people and might pay off in unexpected ways. I’m very grateful to citizens of the Garden State for their generous support! ↩
I want more smart people to go into public service, and fewer smart people doing weird, illegal finance things. So, below are two brief examples of good government that I hope might inspire some plucky students to consider becoming bureaucrats.
The Supplemental Nutrition Assistance Program (SNAP), also known as food stamps, is a federal program that gives low-income Americans money for food. It feeds millions of Americans who might otherwise go hungry.
A great part of SNAP is that it’s dirt cheap. It seems untoward, but cost is an important part of welfare programs. How many people need money? How much will we give them? How much will this improve their standards of living? Helping people is good, but there’s a big difference between a billion dollars and a trillion dollars. SNAP comes in at under three percent of the annual federal budget.
We knew that SNAP would be affordable thanks to estimates from the Congressional Budget Office (CBO), whose economists and policy analysts model costs and budgetary effects. CBO models run on information from the Department of Agriculture (USDA), which administers SNAP, the Census Bureau, which collects demographic information, and the Federal Reserve, which monitors economic conditions.
The planning and analysis of SNAP implicates a handful of federal agencies, but the day to day operations are handled by states themselves, which gives us more than fifty agencies and probably thousands of employees. This sounds like a vast bureaucracy, but employees are very cheap! The total cost of all federal employees in 2019 was around 300 billion dollars, which is roughly the cost of SNAP itself.
For a tiny portion of annual spending, the federal government gets an army of dedicated public servants who provide quality data, analysis, and oversight related to of a wide array of programs like SNAP. This is an excellent and often overlooked feature of the federal government.
In 1957, nine black Americans enrolled in Little Rock Central High School. They had been barred from attending by segregation, but the Supreme Court had just ruled that this was unconstitutional1. In a shocking move, Arkansas governor Orval Faubus defied the Supreme Court and ordered the Arkansas National Guard to forcibly block the students from entering school.
The federal government obviously has to intervene here. You can’t have governors using pseudo-armies to defy court orders. It isn’t as feel-good as feeding hungry people, but an important function of government is maintaining order and enforcing laws. Fortunately, President Eisenhower understood this and sent in the Army.
Whenever normal agencies prove inadequate to the task and it becomes necessary for the Executive Branch of the Federal Government to use its powers and authority to uphold Federal Courts, the President’s responsibility is inescapable.
In accordance with that responsibility, I have today issued an Executive Order directing the use of troops under Federal authority to aid in the execution of Federal law at Little Rock, Arkansas. This became necessary when my Proclamation of yesterday was not observed, and the obstruction of justice still continues.
—President Eisenhower, 1957
Eisenhower’s ability to do this requires a lot of moving parts.
First, there needs to be a federal law enforcement mechanism which can force states to do things. That’s a professional military. Second, there needs to be some legitimate authority for the President to use force to restore order. Congress provided this in the Insurrection Act of 1807. Finally, the President needs to have the character to use the tools at his disposal wisely. We can’t have the military shooting pickpockets in New York City.
It’s an old American pastime to complain about Congress and the president, but they are public servants who make important legislative and executive decisions. Had Congress not recognized the importance of a strong military, if they had been more squeamish about enforcing laws, or if Eisenhower had been a little racist himself, school integration could have played out very differently.
Government is necessary, but good government takes good public servants. We need more smart people in every branch, every agency, and every office. Math students tend to not have the charisma necessary to be politicians2, but there are a lot of other good options!
There are openings every summer at the Department of Energy, the NSA always wants mathematicians (but has some questionable hiring practices), and you definitely qualify for jobs in the 1500 series. This includes jobs at Census, IRS, GAO, NIST, Treasury, and more. If you are interested in the judiciary, then you will be comforted to know that math students tend to do pretty well on the LSAT.
I would encourage all math students to at least consider an internship in government. You can spare one summer to try it out!
The Court also upheld segregation years earlier, but American segregation was happening before the Court ever said a word about it. ↩
Though there was Ira Glasser, so you never know. ↩
The basic idea is that we want to prove the identity
\[\begin{equation} \label{goal} \sum_{k = 0}^{n - 1} {6n \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{6n - k} = \sum_{k = 0}^{n - 1} R(k) {6k \choose k} \left( \frac{1}{6} \right)^k \left( \frac{5}{6} \right)^{5k} \end{equation}\]where $R(k)$ is some explicit rational function. The reason is that the sum on the right has no $n$ in it, and after seeing $R(k)$ written out, is obviously monotonic because the terms are positive. The sum on the left contains a bunch of $n$’s, so it isn’t obvious if it gets bigger or smaller as $n$ grows.
First, a quick lesson on WZ theory. A sequence $a(n, k)$ is called hypergeometric provided that $a(n + 1, k) / a(n, k)$ and $a(n, k + 1) / a(n, k)$ are both explicit rational functions in $n$ and $k$. Certain nice hypergeometric sequences have a “partner” $b(n, k)$ which satisfies the identity
\[a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k).\]In this case $(a, b)$ is called a WZ pair. It is simple to algorithmically determine whether any explicit sequence is hypergeometric or part of a WZ pair. Our summand happens to be both.
Say that you have a WZ pair $(a, b)$. That is,
\[\begin{equation} \label{wz} a(n + 1, k) - a(n, k) = b(n, k + 1) - b(n, k). \end{equation}\]Let
\[S(n) = \sum_{k = 0}^{n - 1} a(n, k)\]be the sum that we are interested in. Summing over \eqref{wz} from $k = 0$ to $k = n$ gives
\[S(n + 1) - S(n) - a(n, n) = b(n, n + 1) - b(n, 0).\]If we move $a(n, n)$ to the right-hand side, then the left-hand side is telescoping, so summing again recovers $S(n)$ itself. The end-result will be
\[S(n) = \sum_k (a(k, k) + b(k, k + 1) - b(k, 0)),\]where the bounds are too tedious to determine. In our case it turns out that $b(k, 0) = 0$, and by some standard theory about hypergeometric sequences $b(k, k + 1)$ is a rational multiples of $a(k, k)$, so we can write
\[S(n) = \sum_k R(k) a(k, k),\]the kind of sum we desired.
The proof now is merely a matter of executing the steps outlined in the
previous section. All of these have been implemented in Maple or
Mathematica for decades. If we download
EKHAD and read it
into Maple, then executing zeillim(a(n, k), k, n, N, 0, 1)
produces output
that implies \eqref{goal}. The rational function is
The age distribution of Austrians walking around cities skewed much older than what I would expect in America. It was not uncommon for most people on a tram or sitting at a cafe to be in their forties, fifties, or higher. Young people were around but only made up majorities in the city center or at night. This seems different than what you would see in New York City, Atlanta, DC, or even my small hometown in South Carolina.
It turns out that Austrians are measurably older than Americans. In the US, about 1 in 3 people (36%) are 50 or older. In Austria, the number is about 2 in 5 (42%). In total, Austria’s population is the 14th oldest in the world, while the US is 62nd. This doesn’t completely explain what I saw (there are regional differences in age distributions, plus I was subject to confirmation bias), but it makes me feel a little less crazy.
I’d be concerned if I was Austrian. Aging populations mean fewer workers support more people, so there’s less money to go around for everyone. Austria could try to get more workers, but their fertility rates have fallen to around 1.4 children per woman, the country seems to have a reputation for being unfriendly to immigrants and raising the retirement age makes people angry. They haven’t had much success improving their slow productivity growth, either.
Austria isn’t alone with its demographic problems. Most developed countries are in the same boat, which the New York Times described in an excellent write-up earlier this year. The key takeaways are that Europe and East Asia are aging very fast, and the US is aging slower thanks to high immigration. The only places getting younger in the near future are Africa and India.
This is a great time to make it easier to move to and stay in the US. Far more people want to move here than we accept. Our current immigration policy is to make most of them to wait, possibly for decades. Why not let them in?
The Cato Institute has a collection of proposals for how to increase immigration. One that I like, from the perspective of deal-making, is expanding visa access for high-skilled workers. Republicans don’t want to take the huddled masses yearning to be free, but maybe they’ll agree to take the educated masses yearning to work at Google1.
At a bare minimum, we should get back to normal, pre-COVID levels of immigration. We handed out embarassingly few green cards during the pandemic. Let’s open the doors again.
Did you know that the US has a special tax treaty with India that lets Indian students in the US take the standard deduction? A non-Indian foreign graduate student pays around a thousand dollars a year more in taxes than Indians or Americans. ↩
Take, for example, power generation. Every factory needs power. One very efficient late game technology is nuclear power, which requires the following supply chain:
The catch is that step 2 is probabilistic. For each unit of uranium ore that you process, there is a 99.3% chance to obtain uranium-238 and a 0.7% chance to obtain uranium-235. Let’s call these “bad” uranium and “good” uranium, respectively. Step 3 requires that you have 10 bad uranium and 1 good uranium. These probabilities make the setup for nuclear power very slow.
Things liven up once you generate 40 good uranium. Then, via another late-game technology, you can put 40 good uranium into a machine that spits out 41 good uranium. This is 41 total good uranium—the 40 you put in plus one new one. It doesn’t sounds like much but multiplying your supply by $41 / 40$ forever would lead to exponential growth.
Unfortunately, this is not how things work. The process to turn 40 good uranium into 41 good uranium only works on 40 uranium at a time per machine you have to run it. The first time you run it you will have 41 good uranium, but you can only put exactly 40 back into the process. The lone new uranium sits there, useless.
This brings us to the neat question. If you had infinitely many machines to run this process, then the amount of good uranium at step $n$ is given recursively by
\[\begin{align*} U_0 &= 40 \\ U_{n + 1} &= U_n + \left \lfloor \frac{U_n}{40} \right \rfloor = \left \lfloor \frac{41}{40} U_n \right \rfloor. \end{align*}\]What is the asymptotic value of $U_n$?
A good idea is to ignore the floor and guess that the answer is close to $(41 / 40)^n \cdot 40$ anyway. This is wrong, but it’s not that wrong.
$n$ | $U_n$ | $(41 / 40)^n \cdot 40$ | ratio |
---|---|---|---|
0 | 40 | 40 | 1 |
1 | 41 | 41 | 1 |
2 | 42 | 42 | 1 |
40 | 80 | 107 | 0.745 |
100 | 291 | 472 | 0.616 |
Like any good question, you can find a version of this in the OEIS. If, instead of $41 / 40$, you use $3 / 2$, then you will find A061418. This led me to a paper of Odlyzko and Wilf studying similar sequences.
Adapting the ideas in Wilf’s paper, it turns out that $U_n$ is exponential but smaller than $(41 / 40)^n \cdot 40$. There is a constant $c$ such that
\[U_n \sim (41 / 40)^n 40 c \quad ; \quad c = 0.574256\dots\]So you lose about 43% of your exponential capacity to waiting, but this is still pretty good!
The constant $c$ does not seem to be expressible in terms of anything well-known. There is a “formula,” but it is useless for practical purposes.
Unsurprisingly, Wilf and Odlyzko were not considering uranium processing in Factorio. They were working on a solution to the Josephus problem.
The Josephus problem is this: Arrange $n$ people in a circle and execute every $q$th person until there is only one left. Where should you stand in the initial circle to be the surviving person? This position is called $J_q(n)$. The $q = 2$ case has the famous solution $J_2(n) = 2(n - 2^{\lfloor \log_2 n \rfloor}) + 1$, but there aren’t closed forms beyond that.
Concrete Mathematics has a neat algorithm to solve this problem. Once you pass over someone, move their index to the smallest index not yet seen. (If $q = 3$ then $1$ will become $n + 1$ and $2$ will become $n + 2$.) The last person to be executed will have the label $qn$, so you just need to determine the original index of the person who is eventually labeled $qn$. The sequences of Wilf and Odlyzko help do this.
Our sequence helps solve a similar Josephus-type problem. Arrange $n$ people in a circle and execute every $q$th person starting from the first one. That is, you will always execute the first person, then count off by $q$ from there. Under the same relabeling scheme as before, the last person to be executed will have label $q(n - 1) + 1$. It turns out that the label of this person after going in “reverse” $k$ steps is $qn - U_k^{(q)}$, where $U_k^{(q)}$ is defined by
\[\begin{align*} U_0^{(q)} &= q - 1 \\ U_k^{(q)} &= \left\lfloor \frac{q}{q - 1} U_{k - 1}^{(q)} \right\rfloor. \end{align*}\]This sequence solves the problem because we can stop computing as soon as $qn - U_k^{(q)} \leq n$. That is, as soon as $U_k^{(q)} \geq (q - 1)n$.
This sequence is also exactly the same form as our uranium sequence if we set $q = 41$. That is, when we are producing good uranium in Factorio, we are implicitly solving this modified Josephus problem where we execute every forty-first person. If you have $U$ good uranium the first time that you cross $40n$, then then the survivor stood at initial position $41n - U$.
This is not quite proving that Factorio is Turing complete (it is), but it is a neat fact to notice. Factorio has more mathematical depth than any game I’ve played before. I hope that this one example demonstrates that.
]]>This year I won two of the TA teaching awards given by the Rutgers math department to graduate students. In the past 25 years the department has given 100 such awards, but only five people have ever won more than one, and none more than two. I don’t know if this is a Joe Pesci, “Thank you” moment, or a Sally Fields, “You like me!” moment. Either way, I am glad to be in the company of many excellent teachers1.
I am particularly happy to contribute to the excellence of teaching in my academic bloodline. I wrote a small script to gather data on the recipients of teaching assistant awards, and am pleased to share the results.
Which advisor at Rutgers produces students who win the most teaching assistant awards? Doron Zeilberger.
Advisor2 | Students | TA Excellence awards |
---|---|---|
Zeilberger | 21 | 11 |
Huang | 12 | 5 |
Woodward | 9 | 4 |
Tunnell | 6 | 4 |
Lepowsky | 13 | 4 |
Of course, Professor Zeilberger has many students, which gives him more chances to get lucky. To account for this, take advisors with at least five students and sort by awards per student. Zeilberger is still second.
Advisor | Students | TA Excellence awards per student |
---|---|---|
Tunnell | 6 | 0.666667 |
Zeilberger | 21 | 0.52381 |
Woodward | 9 | 0.444444 |
Huang | 12 | 0.416667 |
Wilson | 5 | 0.4 |
To really get a feel for this data you should just look at the picture.
While Zeilberger is very far away from the rest of the faculty, this scatter plot does not convey how far away. Almost all advisors “produce” zero TA teaching awards.
Even if you think that the TA Excellence awards are flukes, according to the department only five math graduate students have ever won the more prestigious School of Arts and Sciences Award for Distinguished Contributions to Undergraduate Education. Four of them—Andrew Baxter, Lara Pudwell, Paul Raff, and Matthew Russell—are Zeilberger students.
It does not surprise me that Dr. Z gets the best teachers. Here is the first line of Opinion 77:
We professional academic mathematicians are lucky to be paid for something we enjoy doing the most, mathematical research. Many of us also enjoy teaching, and I for one, would be willing to pay, if I had to, for the privilege to teach, even Freshman calculus.
Whether Dr. Z puts this ethos into his students, or just happens to attract ones that already have it, I don’t know. But there is obviously something in the water, so to speak.
It has been a real joy to teach all of my students over the past four years. Every thankful email I get, every nice (or constructive!) comment in evaluations, every time someone (unwisely) asks me for a letter of recommendation, I feel happy that I’m helping someone along. If these awards reflect anything, they reflect that.
Each award comes with \$200. I’ve donated \$100 to the OEIS and $100 to the Wikimedia foundation. ↩
This data was collected by matching teaching award recipients with their listed advisors on the graduation page. For current students who had not graduated, I attempted to determine their advisor myself. This worked for all but three current students.
The data also only counts students who existed after the creation of the TA Excellence award. For example, Tunnel had 7 students in total, but one was before the award existed. ↩
Rutgers is on strike.
The strike has been brewing around campus for the past few months. Our union and the administration have been unable to agree on new contract terms since our previous contract expired last year. So, there’s a strike.
Being good critical thinkers, we know to distrust the statements of both sides. When the union says that public sector strikes in New Jersey are not illegal, they are misleading (lying). When the administration says that there’s a budget deficit, they don’t explain why such deficits have disappeared in the past.
Unsavory behavior aside, the faculty union is the only group that represents graduate students at Rutgers. That’s why I’ve been a dues-paying member since I arrived here, and why I encourage all graduate students to be the same.
But while we’re here, all of this contract talk raises a good question: What should PhD students be paid? Take, for example, a teaching assistant. The ten-month TA contract stipulates that a TA should not work more than 15 hours a week on average. With our \$31,000 salary, that works out to about \$52 per hour. Is this fair?
Well, this calculation leaves out all the other things that graduate students are expected to do. They conduct research, organize seminars, serve on committees, prepare papers, mentor students, and so on. The actual hourly pay is closer to \$20 or \$30 if you include these things. Is this fair?
To add to the confusion, why are we treating degree requirements like paid labor? Medical students, law students, and Master’s students also work very hard on their degrees, and we don’t pay them a dime. In fact, we charge them! Is it fair that PhD students want to include things like “going to class” and “listening to talks” in their billable hours?
It’s difficult to answer this objectively, because the question is more about values than labor. PhD programs produce researchers and professors. To the extent that you value these things and want a more diverse cast of people trying to enter these fields, you should pay PhD students more. Our stipend is a bit arbitrary, probably determined by more by policy than the value of our labor. (Why do we get paid more than adjuncts to teach less than them? Someone decided we should!)
But “our stipends are arbitrary” does not mean that they should be low. Paying smart people relatively small amounts of money to learn hard things or solve difficult problems seems like a great return. Just look at stories about NYU’s now-free medical school. If every medical school were free, then poorer people would be less intimidated by the sticker price. Perhaps more students would enter areas that are important but pay worse, like family medicine. Paying PhD students seems like a good idea that could be replicated in other areas.
As for the exact numbers, I don’t know. I think that graduate students should be able to live modest lives. They should not worry about day-to-day expenses. They should be able to enjoy themselves, but not to an arbitrary degree. For me, our \$30,000 stipend did this when I arrived. But with inflation over the past few years, we are materially worse off than we were before, and a small raise would not change this. I don’t know why we should accept such a deal if we could avoid it.
For this reason, I throw my hat in with the union, warts and all. They are infinitely better equipped to make arguments than I am.
]]>It was posted by Twitter user @lporiginalg, an account with a Roman statue of Socrates profile picture that mostly posts conservative takes and makes fun of liberals. The idea is that the teacher is showing young kids two different ways to parse the pictured expression, and lporiginalg and his followers want to point and laugh because there is only one right answer.
The answer has always been and always will be 1. If you’re arguing against that my follow up question to you would be “What is a woman?” https://t.co/HJ8hgmH1xe
— Fugazi Financial Management (@FrankDanconia6) February 9, 2023
Every once in a while a picture like this is posted and righteous internet users cry that America’s education system is falling apart. They often insinuate that “woke liberals” are making our kids dumber because they declare all answers equally right, even the wrong ones. Normally I roll my eyes at this dance, but this one was particularly annoying.
The way we write math is completely arbitrary. Notation and order of operations is not something that we can “get wrong” the way people imply here. Conventions exist to make our lives easier. For example, I type a lot of my math and end up writing things like $1 / 2(x + 1)$ to mean $\frac{1}{2(x + 1)}$. This is faster to type and saves some keystrokes by avoiding parentheses. Some of my colleagues might think I mean $\frac{x + 1}{2}$, but we would pretty quickly understand what the other meant and adjust accordingly. Neither way is “wrong,” it’s just an ambiguous expression without a convention telling you what it means.
We teach students PEMDAS because because it’s a useful, common convention. It isn’t “The Law of Math, as Ordained by Our Father in Heaven, Hallowed be His Name.” As we see in the picture, PEMDAS doesn’t even resolve every ambiguity unless you also mention working left-to-right.
The important lesson about order of operations is that a student understands they need an order of operations when parsing an expression. Then they should understand how PEMDAS+left-to-right parsing is normally used and how to spot ambiguities, but the conceptual understanding is more important than any algorithm we give them. To the extent that this picture illustrates that, I think it reflects well on the lesson. To the extent that people use this picture to heckle a teacher, I think the whole affair reflects poorly on the hecklers.
]]>This implies that the series $\sum_{k = 1}^\infty \frac{1}{F_k}$ converges because it is essentially a geometric series with ratio $\phi^{-1}$. In particular, this means that the tails of this series must go to zero, which means that the reciprocals of the tails get big. More explicitly, the sequence
\[S(n) = \left(\sum_{k = n}^\infty \frac{1}{F_k}\right)^{-1}\]goes to infinity. Here are the first few terms:
n | S(n) |
---|---|
2 | 0.4237 |
3 | 0.7354 |
4 | 1.1629 |
5 | 1.8991 |
6 | 3.0623 |
7 | 4.9615 |
8 | 8.0238 |
Do you notice anything interesting about this table? Let’s see it one more time, with a new column.
n | S(n) | round(S(n)) |
---|---|---|
2 | 0.42404 | 0 |
3 | 0.73537 | 1 |
4 | 1.1629 | 1 |
5 | 1.8991 | 2 |
6 | 3.0623 | 3 |
7 | 4.9615 | 5 |
8 | 8.0238 | 8 |
If you round the tails, you get the Fibonacci numbers again! That is, it looks like
\[\mathrm{round}\left(\left(\sum_{k = n}^\infty \frac{1}{F_k}\right)^{-1}\right) = F_{n - 2}.\]This turns out to be correct. In fact, it turns out to be correct for a pretty broad class of sequences. Here, I’m going to show a quick proof, summarize what we know about these sequences, and point out where there are gaps in our knowledge.
The intuition here is that, since $F_n$ is basically $\phi^n / \sqrt{5}$, something like this should be true:
\[\sum_{k \geq n} \frac{1}{F_k} \approx \sum_{k \geq n} \frac{\sqrt{5}}{\phi^k} = \frac{\sqrt{5}}{\phi^{n - 1} (\phi - 1)}.\]Then we reciprocate both sides, and get something useful to play with. The proof just makes this intuition more rigorous. It uses the asymptotic identity $(1 + O(a(n)))^{-1} = 1 + O(a(n))$, valid whenever $a(n) \to 0$.
Proof. By Binet’s formula, $F_n = A \phi^n + B \psi^n$ for some constants $A$ and $B$, where $\phi$ and $\psi$ are the golden ratio and its conjugate, respectively. Therefore,
\[\begin{align*} \frac{1}{F_k} &= (A \phi^k)^{-1} (1 + O((\psi / \phi)^k))^{-1} \\ &= (A \phi^k)^{-1} (1 + O((\psi / \phi)^k)) \\ &= (A \phi^k)^{-1} + O((\psi / \phi^2)^k). \end{align*}\]If we sum this over $k \geq n$, then we obtain
\[\sum_{k \geq n} \frac{1}{F_k} = (A \phi^{n - 1} (\phi - 1))^{-1} + O((\psi / \phi^2)^n).\]Reciprocating yields
\[\begin{align*} \left(\sum_{k \geq n} \frac{1}{F_k}\right)^{-1} &= ((A \phi^{n - 1} (\phi - 1))^{-1} + O((\psi / \phi^2)^n))^{-1} \\ &= A \phi^{n - 1} (\phi - 1) (1 + O((\psi / \phi)^n)) \\ &= A \phi^{n - 1} (\phi - 1) + O(\psi^n). \end{align*}\]Now, note that $A \phi^{n - 1} (\phi - 1) = F_n - F_{n - 1} + O(\psi^n)$, and that $F_n - F_{n - 1} = F_{n - 2}$. This gives us
\[\left(\sum_{k \geq n} \frac{1}{F_k}\right)^{-1} = F_{n - 2} + O(\psi^n).\]Since $|\psi| < 1$, eventually the left-hand side is within $1/2$ of $F_{n - 2}$, so rounding the left-hand side for sufficiently large $n$ gives $F_{n - 2}$. $\blacksquare$
The key insights were something like this:
It turns out that this proof applies to nearly any sequence whose characteristic polynomial satisfies these properties. The theorem goes something like this.
Theorem. Let $c(n)$ be an integer-valued sequence which satisfies a linear recurrence with constant coefficients. If the characteristic equation of this recurrence has one root outside the unit circle and all other roots strictly inside the unit circle, then
\[\mathrm{round} \left( \left( \sum_{k \geq n} \frac{1}{c(k)} \right)^{-1} \right) = c(n) - c(n - 1)\]for sufficiently large $n$.
Fix an integer-valued sequence $c(n)$ which satisfies a linear recurrence with characteristic polynomial $P(x)$. Let
\[S(n) = \left( \sum_{k \geq n} \frac{1}{c(k)} \right)^{-1}.\]The question is what happens to $\mathrm{round}(S(n))$.
If the largest root of $P(x)$ is outside the unit circle and the smaller roots are inside the unit circle, then $S(n)$ rounds to $c(n) - c(n - 1)$ for sufficiently large $n$. This is the previous theorem I mentioned.
There are lots of papers which say this, then provide an example of its application. They mostly have the flavor of, “look at an infinite family of polynomials I found which satisfies this property that we need.”
When does $S(n)$ not round to $c(n) - c(n - 1)$? Is the previous condition about the roots necessary? What if there are roots on the unit circle?
When does $\mathrm{round}(S(n))$ satisfy a linear recurrence with constant coefficients? Perhaps it does not equal $c(n) - c(n - 1)$, but it might still satisfy a recurrence of some kind. This would be a neat closure property of C-finite sequences.
These last two questions seem open.
Let me demonstrate that “one root is outside the unit circle and the others are inside” is not a necessary condition. Take the example sequence $c(n) = 2^n - 1$, which has the characteristic polynomial $(x - 1)(x - 2)$. Let
\[S(n) = \left( \sum_{k \geq n} \frac{1}{2^k - 1} \right)^{-1}.\]There is very convincing data that $S(n)$ will round to $2^{n - 1}$:
n | S(n) |
---|---|
2 | 0.622 |
3 | 1.648 |
4 | 3.658 |
5 | 7.662 |
6 | 15.665 |
7 | 31.666 |
However, the previous proof does not work. If we try to follow the argument again, we’ll get
\[S(n) = 2^{n - 1} + O(1),\]which is not good enough to say what $S(n)$ will round to. This is because one root of the polynomial is on the unit circle rather than inside it.
Nevertheless, there is an argument. We just need to be more careful with our asymptotics. Start by writing
\[\begin{align*} \frac{1}{2^k - 1} &= \frac{1}{2^k} \left(1 + \frac{2^{-k}}{1 - 2^{-k}}\right) \\ &= \frac{1}{2^k} + \frac{2^{-2k}}{1 - 2^{-k}}. \end{align*}\]If we sum this over $k \geq n$, then we get
\[S(n)^{-1} = 2^{-n + 1} + \sum_{k \geq n} \frac{2^{-2k}}{1 - 2^{-k}}.\]Call that last sum $E(n)$. In the previous proof, we essentially say $E(n) = O(2^{-2n})$ and call it a day, which ends up giving us an $O(1)$ error. By being more careful, we can determine the constant on $2^{-2n}$ and thereby improve our final error term. It looks like this:
\[\begin{align*} \sum_{k \geq n} \frac{2^{-2k}}{1 - 2^{-k}} &= \sum_{k \geq n} (2^{-2k} + O(2^{-3k})) \\ &= \frac{2^{-2n}}{1 - 2^{-2}} + O(2^{-3n}). \end{align*}\]So, in fact, $E(n) = \frac{4}{3} 2^{-2n} + O(2^{-3n})$. If we stick with $E(n)$, we have
\[\begin{align*} S(n) &= (2^{-n + 1} + E(n))^{-1} \\ &= 2^{n - 1} (1 + 2^{n - 1} E(n))^{-1} \\ &= 2^{n - 1} (1 - 2^{n - 1} E(n) + O(2^{2n} E(n)^2)). \end{align*}\]Now we can plug-in our improved estimate for $E(n)$:
\[\begin{align*} S(n) &= 2^{n - 1} - 2^{2n - 2} (2^{-2n} (1 - 2^{-2})^{-1} + O(2^{-3n})) + O(2^{-n}) \\ &= 2^{n - 1} - \frac{1}{3} + O(2^{-n}). \end{align*}\]The sequence $S(n)$ is almost exactly a third away from $2^{n - 1}$, so eventually we’ll have $\mathrm{round}(S(n)) = 2^{n - 1}$.
The takeaway is that sometimes $S(n)$ will round to $c(n) - c(n - 1)$ even when the characteristic equation of $c(n)$ has a root on the unit circle. So we still don’t know a necessary condition for this rounding behavior.
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